• Title: Integrable Functions

  • Series: Fourier Transform

  • Chapter: Fourier Series

  • YouTube-Title: Fourier Transform 5 | Integrable Functions

  • Bright video: https://youtu.be/2NUNRQEp7YE

  • Dark video: https://youtu.be/LgRBO95N0xM

  • Quiz: Test your knowledge

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  • Subtitle on GitHub: ft05_sub_eng.srt missing

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  • Quiz Content

    Q1: Consider the function $f: [0,1] \rightarrow \mathbb{R}$ given by $f(0) = 0$ and $f(x) = \frac{1}{x}$ for $x > 0$. Is the function integrable?

    A1: No, because $\int_0^1 | f(x) | , dx = \infty $.

    A2: Yes, because $\int_0^1 | f(x) | , dx < \infty $.

    A3: No, because $f$ is unbounded.

    A4: Yes, because $f$ is bounded.

    Q2: Consider the function $f: [0,1] \rightarrow \mathbb{R}$ given by $f(0) = 0$ and $f(x) = \frac{1}{\sqrt{x}}$ for $x > 0$. Is the function integrable?

    A1: No, because $\int_0^1 | f(x) | , dx = \infty $.

    A2: Yes, because $\int_0^1 | f(x) | , dx < \infty $.

    A3: No, because $f$ is unbounded.

    A4: Yes, because $f$ is bounded.

    Q3: Consider the function $f: [0,1] \rightarrow \mathbb{R}$ given by $f(0) = 0$ and $f(x) = \frac{1}{\sqrt{x}}$ for $x > 0$. Is the function square-integrable?

    A1: No, because $\int_0^1 | f(x) |^2 , dx = \infty $.

    A2: Yes, because $\int_0^1 | f(x) |^2 , dx < \infty $.

    A3: No, because $f$ is unbounded.

    A4: Yes, because $f$ is bounded.

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